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Let $i\colon \mathbb{R}P^2\to \mathbb{C}P^2$ the usual embedding $i( [x:y:z] )=[x:y:z]$. I want to compute the induced map by $i$ in the second homology groups with $\mathbb{Z}/2$ coefficients. I think the easiest way should be to use the intersection pairing. Let $$S=\{ [z_0:z_1:z_2]\in \mathbb{C}P^2\mid a_0z_0+a_1z_1+a_2z_2\}$$ be a $\mathbb{C}P^1$ in $\mathbb{C}P^2$, and $$N=\{ [x_0:x_1:x_2]\in \mathbb{C}P^2\mid x_i\in\mathbb{R}\}.$$ If I show that $N\cdot S=|N\cap S| \ne 0$ for some $(a_1,a_2,a_3)\in \mathbb{C}^3$, then in particular the induced map will be an isomorphism $\mathbb{Z}/2\to\mathbb{Z}/2$.

However, I think that $N$ and $S$ must intersect transversally to have the equality $N\cdot S=|N\cap S| $. How can I do this? I am not familiar with the intersection pairing, so any suggestions, etc will be welcome.

Laszlo
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  • The map $F_t:\mathbb{R}P^2\rightarrow \mathbb{C}P^2$, $[a,b,c]\mapsto [a+(1-t)ib,(1-t)b+itc,(1-t)c]$ is an isotopy of $i$ to the embedding $j:[a,b,c]\mapsto [a+ib,ic,0]$. Let $N'$ be the image of $j$ in $\mathbb{C}P^2$. Take $a_0=1$, $a_1=a_2=0$. Since the intersection number depends only on the isotopy class we get $N\cdot S=N'\cdot S=|N'\cap S|=|{[0,1,0]}|=1$. – Tyrone May 29 '18 at 09:57

1 Answers1

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I like this strategy, though maybe just take $a_0 = 0$, $a_1 = 1$, $a_2 = -i$ so that we are intersecting $[x_0 : x_1 : x_2]$ with $[z_0 : z_1 : iz_1]$. The only way to scale the last two coordinates to both be real is if $z_1 = 0$; in that case we are left with the points $[z_0: 0 : 0]$. Up to scalar change, there is only one of these, $[1 : 0 : 0]$. This is of course real.

We should now check that the intersection at $[1:0:0]$ is transversal. I would write $T_{[1:0:0]} \Bbb{CP}^2$ as $\Bbb C^2$, generated by the tangent vectors changing the second or third coordinate by some complex number. $T\Bbb{RP}^2$ is $\Bbb R^2 \subset \Bbb C^2$, the real elements; the complex line $\Bbb{CP}^1$ has tangent space given by the complex subspace $\Bbb C \subset \Bbb C^2$ generated by $(1,i)$. These two vector spaces are in direct sum: their intersection is zero (you of course cannot write $(\lambda, i\lambda)$ for two reals unless $\lambda = 0$), and they sum to the dimension of the total space. So this is a transversal intersection, as desired, and the intersection product is indeed 1.